Three-digit numbers are 999-100+1 = 900.

For any arrangement of three numbers, there is only one in which the numbers are arranged so that the units are less than the tens, which are less than the hundreds: 10x9x8 = 720 are the total different combinations of three different integers. Only one of the six possible arrangements (3! = 6) matches up with the restraints in the problem. Thus, 720/6 = 120.

Then, 120/900, which simplifies to 2/15. Answer C

The total 3 digit numbers can be formed by FCP i.e = 9*10*10 (we have digit from 0 -9 and there are no restriction then the hundred digit can be from 1-9,as it cannot start from 0 , tens digit can be 0-9 and units can also be 0-9).

Now we have to choose the three digit numbers in descending order -

We can leave out 100 - 200 = as they donot form in descending order

Next from 200 to 200 = we have only one number =210

Next from 300 to 400 = we have 3 nos = 310,320,321

Next from 400 to 500 = we have 6 nos = 410,420,421,430,431,432

Now we find that it is following a pattern and i.e



200 = 1

300 = 3 (2+1 from previous possible ways , 2 is the hundred unit )

400 = 6 (3+3)

500 = 10 (4+6,similarly 4 is the hundred digit and 6 in the value from the previous)

600 = 15(5+10)

700 = 21 (6+15)

800 = 28 (7+21)

900 = 36(8+28)

Therefore no. of possible ways = 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 =120

Therefore probability = \(\frac{120}{900} = \frac{2}{15}\)
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We see that the hundreds digit can be any digit from 2 to 9.

If we start with the hundreds digits as 2, then the tens digit must be 1, and the units digit must be 0. We have 1 descending number when the hundreds digit is 2.

If the hundreds digits is 3, then if the tens digit is 2, the units digit can be1 or 0; if the tens digit is 1, the units digit must be 0. We have 3 descending numbers when the hundreds digit is 3.

Now, if the hundreds digit is 4, then if the tens digit is 3, the units digit can be 2, 1, or 0; if the tens digit is 2, the units digit can be1 or 0; if the tens digit is 1, the units digit must be 0. We have 6 descending numbers when the hundreds digit is 4.

From this point, we can see a pattern:

hundreds digit = 2 → number of descending numbers = 1

hundreds digit = 3 → number of descending numbers = 1 + 2 = 3

hundreds digit = 4 → number of descending numbers = 1 + 2 + 3 = 6

So the number of descending numbers when the hundreds digit is 5 is 10, 6 is 15, 7 is 21, 8 is 28, and 9 is 36. Therefore, there are a total of 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 = 120 three-digit descending numbers.

Since there are 999 - 100 + 1 = 900 three-digit numbers, the probability of picking a descending number is 120/900 = 12/90 = 2/15.

Answer: C
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